In the previous post Solution of First Order Linear Differential Equations, we discussed methods for finding solutions to differential equations of the form:
\[\frac{dx}{dt}+p(t)x=q(t) % Equation (1)\]This time, we will explore the solution to one of the slightly modified nonlinear differential equations, the Bernoulli differential equation.
Form of the Bernoulli differential equation
The form of the Bernoulli differential equation is as follows:
\[\frac{dx}{dt}+p(t)x=q(t)x^n % Equation (2)\]Here, let us assume that $p(t)$ and $q(t)$ are continuous functions in the interval we are analyzing, and $n$ is a real number.
If $n=0$ or $n=1$, it becomes a linear differential equation, so we can say that we are interested in the case where $n$ is not 0 or 1 in the Bernoulli differential equation.
Solution to the Bernoulli differential equation
The key to the solution of the Bernoulli differential equation is to transform the nonlinear equation in Equation (2) into a linear form.
To do this, let’s divide both sides of Equation (2) by $x^n$.
\[\Rightarrow x^{-n}\frac{dx}{dt}+p(t)x^{1-n}=q(t) % Equation (3)\]Now, let’s think of replacing $x^{1-n}$, which is on the left-hand side of Equation (3), with a new function $u=x^{1-n}$ to make it similar to Equation (1).
One thing we can confirm here is the result of differentiating $u$ with respect to $t$, which is obtained by applying the chain rule to the right-hand side of $u=x^{1-n}$:
\[\frac{du}{dt}=(1-n)x^{-n}\frac{dx}{dt} % Equation (4)\]Therefore, Equation (3) can be written as follows:
\[Equation (3)\Rightarrow \frac{1}{1-n}\frac{du}{dt}+p(t)u=q(t) % Equation (5)\]In other words, by appropriately applying substitutions to the Bernoulli differential equation, it can be transformed into a general first-order linear differential equation.
Example
Solving example problems, like any other differential equations, seems to be helpful in understanding how to solve Bernoulli differential equations.
Problem 1
Find the solution to the following differential equation:
\[\frac{dx}{dt}+\frac{1}{t}x=t^2x^2 % Equation (6)\]Solution
The given equation is a Bernoulli differential equation with $n=2$. Dividing both sides of Equation (6) by $x^2$, we get:
\[\frac{1}{x^2}\frac{dx}{dt}+\frac{1}{t}\frac{1}{x}=t^2 % Equation (7)\]Let $u=x^{-1}$. Then,
\[\frac{du}{dt}=\frac{d}{dt}(x^{-1})\frac{dx}{dt}=-\frac{1}{x^2}\frac{dx}{dt} % Equation (8)\]Thus, Equation (7) can be written as:
\[Equation(7) \Rightarrow -\frac{du}{dt}+\frac{1}{t}u=t^2 % Equation (9)\]Multiplying both sides of Equation (9) by $(-1)$, we get:
\[\frac{du}{dt}-\frac{1}{t}u=-t^2 % Equation (10)\]Using the method of finding the solution to a first-order linear differential equation, multiply both sides of Equation (10) by the integrating factor $e^{\int -\frac{1}{t}dt}$.
Since $\int -\frac{1}{t}dt=-\ln t$, $e^{-\ln t}=e^{\ln 1/t}=1/t$ is the integrating factor.
Multiplying both sides of Equation (10) by $1/t$, we get:
\[\frac{1}{t}\frac{du}{dt}-\frac{1}{t^2}u=-t % Equation (11)\]Equation (11) can be rewritten as:
\[Equation(11)\Rightarrow \frac{d}{dt}\left(\frac{u}{t}\right)=-t\] \[\Rightarrow \frac{u}{t} = -\int t dt + C=-\frac{1}{2}t^2+C\] \[\therefore u(t) = -\frac{1}{2}t^3+Ct\]Originally, as $u=1/x$, we can get the following.
\[\therefore x(t) = \frac{1}{-0.5t^3+Ct}\]